EDIT: $R_n$ is the right endpoint of the Riemann sum

Calculate $R_n$ for the function $$f(x) = -\frac{x^2}4-5$$ on interval [0,2]

What I've done so far

$ \Delta x = \frac 2n$ and $f(x_i^*) = \frac{2i}n $

Which leads to $$ \sum_{i=1}^n f(x_i^*)\ \Delta x$$

And once you put the numbers in

$$ \sum_{i=1}^n\left( -\frac{\left(\frac{2i}n\right)^2}4-5\right) \cdot \frac 2n $$

Which leads to my issue of expanding and getting an $i$ or $i^2$ alone.

  • $\begingroup$ @T.Bongers Just added that in the question. $\endgroup$
    – impact_sv1
    Jan 17, 2016 at 21:43

1 Answer 1


You have $$ \sum_{i=1}^n f(x_i^*)\ \Delta x=-\frac {2}{n^3} \sum_{i=1}^n i^2- \frac {10}{n}\sum_{i=1}^n 1 $$ then use $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ which is proved here to get, as $n \to \infty$, $$ \sum_{i=1}^n f(x_i^*)\ \Delta x \to -\frac{32}3 $$

  • 1
    $\begingroup$ Hey thanks dude! Appreciate that! $\endgroup$
    – impact_sv1
    Jan 17, 2016 at 22:03
  • $\begingroup$ You are welcome. @impact_sv1 $\endgroup$ Jan 17, 2016 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.